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Is it true that

$f(t)\star g(-t)$

is not the same as

$f(t)\star h(t)$

if $h(t)=g(-t)$?

($\star$ means cross-correlation).

The example I was thinking of was if $g(t)=t$. Then $(f(t)\star g(-t))(t)=\int_{-\infty}^\infty f(\tau)(-t+\tau)d\tau$, but $(f(t)\star h(t))(t)=\int_{-\infty}^\infty f(\tau)(-(t+\tau))d\tau=\int_{-\infty}^\infty f(\tau)(-t-\tau)d\tau$.

If it's true then wouldn't something like $t^2\star -t$ be ambiguous?

I think I'm most likely making a mistake here, so any help would be greatly appreciated.

John
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1 Answers1

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Something like $t^2\star -t$ is understandable. Since you didn't define $\star$, I'm not sure if or where you made a mistake.

The $f(t)*g(-t)$ is convenient notation and is very common in engineering texts. It's convenient, because it's easier to write $f(t)*g(-t)$ than to write this with more conventional function notation: $$(f*h)(t),$$ where $h(t) = g(-t)$, and be forced to define a new function $h$.

The downside to convenience is that it mixes together the function's "name" and "argument", and can lead to awkward expressions like $(f(t)*g(-t))(t)$ if you want to evaluate this at some point $t$.

b yen
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  • Thanks b yen. In this case, would $(f\star h)(t)$ be a different function (flipped about the y axis) from $(f\star g(-t))(t)$ though? – John May 15 '15 at 01:20
  • If you define $h(t) = g(-t)$, then $(f\star h)(t) = (f\star g(-t))(t)$ are the same function. If you write down your definition of $\star$, it would easier to understand what your wrote in your question. – b yen May 15 '15 at 03:14
  • Hi, I think I was just confused because wikipedia said: "the convolution $fg$ (equivalent to the cross-correlation of f(t) and g(−t) ) ". I interpreted this as $fg = f\star g(-t)$, but it should really be $f*g = g(-t)\star f$, right? – John May 15 '15 at 17:06